Homework Chapter 7

Section 3

Q1

Sampling distribution of the sampel mean:

\(\bar x = 10\) \(SE = \sqrt{\sigma^2/n} = 3/6 = 1/2\)

Q3

\(\bar x = 120\) \(SE = \sqrt{\sigma^2/n} = 1/\sqrt 8 = .353\)

Q15

\(\bar x = 52\) \(SE = \sqrt{\sigma^2/n} = 21/7= 3\)

Q18

\[ \begin{aligned} P(\bar x > 55) & \\ =& P(z > \frac{55 - 52}{3}) \\ =& P(z > 1) = 15.9\% \end{aligned} \]

Q22

\(n = 25\) \(SE = \sqrt{\sigma^2/n} = 16/5= 3.2\)

\[ \begin{aligned} P(\bar x < 69.7) & \\ =& P(z < \frac{69.7 - 74}{3.2}) \\ =& P(z < -1.344) = 8.9\% \end{aligned} \]

Q32

\(\mu = 98.6\), \(\sigma = .8\), \(n = 130\) \(SE = \sqrt{\sigma^2/n} = .8/\sqrt{130}= 0.07016464\)

\[ \begin{aligned} P(\bar x < 98.25) & \\ =& P(z < \frac{98.25 - 98.6}{.0702}) \\ =& P(z < -4.985755) \approx 0\% \end{aligned} \]

2. It’s surprising that this values is so unlikely because seeing this temperature in a single person is not unliklely. However, when you average 130 people together it’s very unlikely!